Given a string containing just the characters ‘(‘
and ‘)‘
, find the length of the longest valid (well-formed) parentheses substring.
For "(()"
, the longest valid parentheses substring is "()"
, which has length = 2.
Another example is ")()())"
, where the longest valid parentheses substring is "()()"
, which has length = 4.
这道求最长有效括号比之前那道Valid Parentheses 验证括号难度要大一些,这里我们还是借助栈来求解,需要定义个start变量来记录合法括号串的起始位置,我们遍历字符串,如果遇到左括号,则将当前下标压入栈,如果遇到右括号,如果当前栈为空,则将下一个坐标位置记录到start,如果栈不为空,则将栈顶元素取出,此时若栈为空,则更新结果和i - start + 1中的较大值,否则更新结果和i - 栈顶元素中的较大值,代码如下:
class Solution { public: int longestValidParentheses(string s) { int res = 0, start = 0; stack<int> m; for (int i = 0; i < s.size(); ++i) { if (s[i] == ‘(‘) m.push(i); else if (s[i] == ‘)‘) { if (m.empty()) start = i + 1; else { m.pop(); res = m.empty() ? max(res, i - start + 1) : max(res, i - m.top()); } } } return res; } };
还有一种利用动态规划Dynamic Programming的解法,可参见网友喜刷刷的博客。
时间: 2024-10-10 04:38:02